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My proof. Let $\varepsilon >0$. Pick $\delta=5\varepsilon$.

If $\left| x-4\right| < \delta$ for $x\in\mathbb{R} ^{\geq 0}$ then $\left| \sqrt {x}-2\right| =\dfrac {\left| x-2\right| } {\left| \sqrt {x}+2\right| }=\dfrac {\left| x-2\right| } {\left| \sqrt {x}\right| +2}=\delta/5\leq \dfrac {5\varepsilon } {\varepsilon }=\varepsilon$.

Note that for $0

Can you check my proof?

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    @juniven Does my delta work for this ? – 2017-01-17
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    First, there is a typo: your $|x-4|$ has turned into $|x-2|$. Also, your second-last $=$ should be $\le$. More seriously, this inequality is not in fact true: you have a situation where $a – 2017-01-17
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    @Kahlert It is not. See the answer of DeepSea below. Actually, in your last line, saying $|\sqrt{x}+2|\leq 5$ can do nothing in your proof. – 2017-01-17
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    @Kahlert All you have to do is to look for some constant $C>0$ for which $|\sqrt{x}+2|>C$ and DeepSea showed $C=\sqrt{3}+2$ whenever $|x-4|<1$. This is the usual approach in $\epsilon$-$\delta$ proof of limits – 2017-01-17

2 Answers 2

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For the epsilon that you have, first make delta to be small, and smaller than $1$. So $|x-4| < 1 \implies 4-|x|=|4| - |x| < |x-4| <1\implies |x| > 3\implies \sqrt{x}+2 > \sqrt{3}+2\implies |\sqrt{x}-2| < \dfrac{|x-4|}{\sqrt{3}+2}< \dfrac{\delta}{\sqrt{3}+2}< \epsilon$, if $\delta < \epsilon(\sqrt{3}+2)$. Thus you can figure out how to take the delta. Simply make $0 < \delta < \text{min}(1,\epsilon(\sqrt{3}+2))$.

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    Does my delta work for this ? – 2017-01-17
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    No, because the denominator is less than....go back to check it again or use mine. – 2017-01-17
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    If $\left| x\right| > 3$ then how did you obtain $\sqrt {x}+2>\sqrt {3}+2$? – 2017-01-17
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    $|x| = x$ because $x > 0$ – 2017-01-17
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    Yes, thanks for answer. – 2017-01-17
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Another very similar way that may add some clarity:

Let $\epsilon>0$ be given. We need to find $\delta>0$ such that if $|x-4|<\delta$, then $|\sqrt{x}-2|<\epsilon$. Note that $$ |\sqrt{x}-2|=\left|\sqrt{x}-2\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}\right|=\left|\frac{x-4}{\sqrt{x}+2}\right|=\frac{|x-4|}{\sqrt{x}+2}. $$ To make the above chain of reasoning useful, we may experiment with values of $\delta$ (often starting with $\delta=1$ although sometimes this may prove to be problematic...occasionally, one needs to choose smaller values of delta to begin with, but $\delta=1$ will do just fine here). We let $\delta=1$ and note that $$ |x-4|<1\Leftrightarrow -1